题目:
Given a sorted integer array where the range of elements are in the inclusive range [lower, upper], return its missing ranges.
For example, given [0, 1, 3, 50, 75]
, lower = 0 and upper =
99, return ["2", "4->49", "51->74", "76->99"].
思路:
为简化操作可以给原数组一个新的上界,比给定的upper上界多1,然后维护一个当前可以覆盖的最大范围,如果下一个元素与最大覆盖范围相差2就往结果集合中添加一个数字,如果相差多于2,就添加一个区间,然后更新最大覆盖范围。Leetcode不知什么时候添加了很多corner case的测试用例,导致我原来写的代码出现溢出。为防止溢出,我在下面的代码片段中将所有int数据类型都定义成为了long long类型。
代码:
class Solution {
public:
vector<string> findMissingRanges(vector<int>& nums, int lower, int upper) {
vector<long long> new_nums; // "long long" is used to overcome overflow
long long new_lower = lower, new_upper = upper;
for(int i = 0; i < nums.size(); ++i) {
new_nums.push_back(nums[i]);
}
new_nums.push_back(new_upper + 1);
long long start = new_lower - 1;
vector<string> ret;
for(int i = 0; i < new_nums.size(); ++i) {
if(new_nums[i] - start > 2) {
ret.push_back(to_string(start + 1) + "->" + to_string(new_nums[i] - 1));
}
else if(new_nums[i] - start == 2) {
ret.push_back(to_string(start + 1));
}
start = new_nums[i];
}
return ret;
}
};